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Standard X

Physics

Charles Law

A 1.00 L ball...

Question

A 1.00 L balloon is filled with helium at 1.20 atm. If the balloon is squeezed into a 0.500 L beaker and doesn’t burst, what is the pressure of the helium?

4.5 atm

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1.4 atm

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2.4 atm

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6.5 atm

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Solution

The correct option is C
2.4 atm

According to Boyle's law: $P_{1} V_{1} = P_{2} V_{2}$

Here $P_{1} = 1.2 a t m : V_{1} = 1 L : V_{2} = 0.5 L$

$\Rightarrow P_{2} = \frac{1.2 \times 1}{0.5}$

$\Rightarrow P_{2} = 2.4 L$

Therefore, the volume of gas produced= 2.4 atm

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A 1.00 L balloon is filled with helium at 1.20 atm. If the balloon is squeezed into a 0.500 L beaker and doesn’t burst, what is the pressure of the helium?

The correct option is C 2.4 atm According to Boyle's law: P1V1=P2V2 Here P1=1.2atm : V1=1L : V2=0.5L ⇒P2=1.2×10.5 ⇒P2=2.4 L Therefore, the volume of gas produced= 2.4 atm

The correct option is C
2.4 atm

According to Boyle's law: $P_{1} V_{1} = P_{2} V_{2}$

Here $P_{1} = 1.2 a t m : V_{1} = 1 L : V_{2} = 0.5 L$

$\Rightarrow P_{2} = \frac{1.2 \times 1}{0.5}$

$\Rightarrow P_{2} = 2.4 L$

Therefore, the volume of gas produced= 2.4 atm